Đặt $\sqrt{x} =a ,\sqrt{y}=b ,\sqrt{z}=c,a+b+c=1 $
Ta có $P= \sum \dfrac{ab}{ \sqrt{a^2+b^2+2c^2} } \leq \sum \dfrac{ab}{\sqrt{ \dfrac{(a+c)^2}{2}+\dfrac{(b+c)^2}{2}}}$
$\leq \sum \dfrac{ab}{\sqrt{(a+c)(b+c)}} \leq \sum ab( \dfrac{1}{2(a+c)}+ \dfrac{1}{2(b+c)}) $
$ =\dfrac{a+b+c}{2} = \dfrac{1}{2}$